Question

Derive an expression for the maximum force experienced by a straight conductor of length l, carrying current I and kept in a uniform magnetic field B.

Answer 1

Consider a conductor PQ carrying current I placed in a uniform magnetic field B. The magnetic field is acting along x-axis and conductor is placed along x-axis.

Let

n = no. of electrons per unit volume

A = area of cross-section of conductor

v_d = drift velocity of electrons

∴ I = n A e v_d ............(1)

Since each electron is moving in uniform magnetic field, therefore, each electron will experience Lorentz force.

Let l be length of conductor PQ

∴ No. of electrons in the conductor = nA.l

Force acting on each electron = \(-(\vec v_d \times \vec B)\)

Force experienced by electrons in the conductor

(∵ the direction of  \(\vec l\)is in the direction of I i.e. from P to Q whereas the direction of  \(\vec v_d\)  is from Q to P. So \(|\vec v_d |= -\vec l vd\) 

or \(\vec F\) = n Ae v_d \((\vec l\times \vec B)\)

Using Eq. (1), we get

\(\vec F = I\ (\vec l \times \vec B)\) .............(2)

In magnitude    

F = IlB sin θ

or F = BIl sin θ .............(3)

Special Cases

(i) When the conductor is held parallel to the magnetic field

i.e. θ = 0°

So F = BIl sin 0° = 0

i.e. if the current carrying conductor is placed parallel to the magnetic field it will experience no force.

(ii) If the conductor is held perpendicular to the magnetic field

i.e. θ = 90°

∴F = BIl sin 90° = BIl

i.e. if the current carrying conductor is placed perpendicular to the magnetic field, it will experience maximum force.